Population Allele Frequency and Heterozygotes in H-W Equilibrium

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In a population with two alleles, B and b, the allele frequency of B is 0.73, indicating that p = 0.73 and q = 0.27. According to the Hardy-Weinberg equilibrium equations, p + q = 1 and p^2 + 2pq + q^2 = 1, the heterozygote frequency can be calculated using 2pq. Substituting the values, the heterozygote frequency is determined by the equation 2(0.73)(0.27). The discussion emphasizes understanding the application of Hardy-Weinberg principles to calculate allele and genotype frequencies in a population.
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A population has two alleles, B and b. the allele frequency of B is 0.73. what is the heterozygotes frequency, if the population is in H-W equilibrium?
 
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How about writing the appropriate equation for HARDY-WEINBERG equilibrium.

Look at

http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/H/Hardy_Weinberg.html
 
two major important equations for Hardy-Weinberg:
p + q = 1
and p^2 + 2pq + q^2 = 1

You know 'p' in your problem: it is .73 Therefore you must know that 'q' is .27
since you know p = .73 and q = .27 you can find 2pq (which is the heterozygote frequency)
{Moderator edit: remainder of solution deleted. The student can solve from here on their own.}[/color]
 
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